$h(t) = t^{2}-4t-4(g(t))$ $g(t) = -2t^{2}-2t+3(f(t))$ $f(n) = n^{2}-6n$ $ h(f(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = (-1)^{2}+(-6)(-1)$ $f(-1) = 7$ Now we know that $f(-1) = 7$ . Let's solve for $h(f(-1))$ , which is $h(7)$ $h(7) = 7^{2}+(-4)(7)-4(g(7))$ To solve for the value of $h$ , we need to solve for the value of $g(7)$ $g(7) = -2(7^{2})+(-2)(7)+3(f(7))$ To solve for the value of $g$ , we need to solve for the value of $f(7)$ $f(7) = 7^{2}+(-6)(7)$ $f(7) = 7$ That means $g(7) = -2(7^{2})+(-2)(7)+(3)(7)$ $g(7) = -91$ That means $h(7) = 7^{2}+(-4)(7)+(-4)(-91)$ $h(7) = 385$